Hi everyone, I got a problem when proving lemmas for some combinatorial problems, and it is a question about integers.

Let

$\sum_{k=1}^m a_k^t = \sum_{k=1}^n b_k^t$

be an equation, where $m, n, t, a_i, b_i$ are positive integers, and $a_i \neq a_j$ for all $i, j$, $b_i \neq b_j$ for all $i, j$, $a_i \neq b_j$ for all $i, j$.

Does the equality have no solutions?

For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem, and even for $n = m$, we have solutions like

$1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$.

For $t > 2$, similar equalities hold:

$1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$ and $1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$,

and we can extend this trick to all $t > 2$.

The question is, if we introduce one more restriction, that is, $|a_i - a_j| \geq 2$ and $|b_i - b_j| \geq 2$ for all $i, j$, is it still possible to find solutions for the equation?

For $t = 2$ we can combine two Pythagorean triples, say,

$5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$,

but how about the cases for $t > 2$ and $n = m$?